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Solve the system of equations :

(EDIT : The problem does not say anything about the nature of $x$ and $y$ (integer, natural number ,..etc.) )

$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3} $

$2x + {\frac {1} {x+y}} = \frac {13} {3}$

I do not know how to approach these types of problems. I tried finding value of $\frac {1} {x+y}$ in terms of $x$ and $y$, but it complicates the problem even more.

Can anyone provide a pointer to what should be done ?

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    $\begingroup$ Just checking: the first term in the second equation is $x$ and not $xy$ $\endgroup$
    – Anurag A
    Commented Jan 12, 2017 at 1:12
  • $\begingroup$ Also, does the problem specify any restrictions on the type of values for x,y? For example, are x,y required to be integers? $\endgroup$
    – quasi
    Commented Jan 12, 2017 at 1:27
  • $\begingroup$ There are no "restrictions" as such mentioned in the problem. It just asks you to "solve" the system of equations... $\endgroup$
    – user399078
    Commented Jan 12, 2017 at 1:29
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    $\begingroup$ @QUANTUM: What do you mean by "it is $4xy$ only"? The question Anurag A asked was whether the first term in the second equation should be $2x$ or $2xy$? $\endgroup$
    – quasi
    Commented Jan 12, 2017 at 1:32
  • $\begingroup$ @QUANTUM: Can you tell us the source of the problem? Is it from a book? $\endgroup$
    – quasi
    Commented Jan 12, 2017 at 1:34

2 Answers 2

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Perhaps this can help: Let $x+y=v$ and $x-y=w$, then the equations can be written as \begin{align*} 3\left(v^2+\frac{1}{v^2}\right)+w^2 & = \frac{85}{3}\\ \left(v+\frac{1}{v}\right)+w & = \frac{13}{3}. \end{align*} Now let $v+\frac{1}{v}=t$, then the system can be rewritten as \begin{align*} 3t^2+w^2 & = \frac{85}{3}+6\\ t+w & = \frac{13}{3}. \end{align*} Now solve for $t$ and $w$...

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  • $\begingroup$ Very nice. It should be $\frac {85}{3}-6$ though, should it not? $\endgroup$
    – Ennar
    Commented Jan 12, 2017 at 1:54
  • $\begingroup$ Yes, very nice. The answers for x,y are ugly, but at least they're symbolic. $\endgroup$
    – quasi
    Commented Jan 12, 2017 at 1:57
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    $\begingroup$ @quasi, real solutions are $x = 2/3, y = -1/3$, $x = 2$, $y = 1$, not that ugly. $\endgroup$
    – Ennar
    Commented Jan 12, 2017 at 1:58
  • $\begingroup$ @Ennar -- I must have made an algebraic error -- will check. $\endgroup$
    – quasi
    Commented Jan 12, 2017 at 2:00
  • $\begingroup$ @Ennar -- yes, those solutions work, my mistake. But there's also a non-real, complex solution which, though symbolic, is a little bit ugly. $\endgroup$
    – quasi
    Commented Jan 12, 2017 at 2:05
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Less elegant than Anurag A's answer and using brute force.

Considering the equations $$4xy + 4(x^2 + y^2) + {\frac {3} { (x+y) ^ 2 } } = \frac {85} {3}\tag 1$$ $$2x + {\frac {1} {x+y}} = \frac {13} {3}\tag 2$$ extract $y$ from $(2)$; this gives $$y=\frac{-6 x^2+13 x-3}{6 x-13}\tag 3$$ Replace $(3)$ in $(1)$ an simplify to get $$\frac{8 (x-2) (3 x-2) \left(24 x^2-118 x+149\right)}{(13-6 x)^2}=0\tag 4$$ and the quadratic term does not show real solution (so,two real roots and two complex conjugate roots for $x$).

The solutions of $(4)$ are simple; when you have them, go back to $(3)$ for the corresponding $y$'s.

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